291217049 1/21/2004 1/2 Richardson’s Extrapolation BLI Integration. Form three integral estimates using every point, every second point, and every third point out of the xi and fi vectors returned by BLI. Ignore the possibility of using the 1/N4 Abli.zip End point trap rule.docx Equation 9 is f xi f xi 1 h2 h4 G1 x c1 G x f ' end f ' beg f '''(endN) 2 f '''(beg ) beg f x dx h 2 12 720 i 1 4c1 O h 6 f v (end ) f v (beg ) G x G 2 x 2 Eqn 1 N 9 c1 Approximate the fifth derivative by G x G 3 x N2 f V x 5 4 f ''' x N end And assume that the coefficient of 1/720 remains the same. Multiply the top of equation 18 by 4 and subtract the second Then the integral to xj can be written as Eqn 2 3G x 4G1 x G 2 x ; G x 4G1 x G 2 x / 3 the top by 9 and subtract the third f x j 1 f x j x j x j 1Multiply g x j g x j 1 x j x j 1 f ' x j f ' x j 1 2 12 G x 9G1 x G3 x / 8 Eqn 3 4 6 Multiply the middle one by 9 and the bottom by 4 x j x j 1 x j x j 1 f ''' x j f ''' x j 1 f ''' and x j subtract f ''' x j 1 720 36 G x 9G 2 x 4G3( x) / 5 Eqn 4 (1) 2 This leads to Figure 1 Intermediate values found by BLI Derivatives at J-1 aredetermined by the curve through J-3to J+1. Derivatives at J are determined by the curve through J-2 to J+2. The four values of f(xj) most relevant for the integral from xj-1 to xj are shown in figure 1. A Lagrange polynomial through the appropriate five points gives third derivatives at xj-1 and xj. ..\Derivatives\Lagrange.docx The goal is to spread N points from x0 to xN such that the error term in the integral is minimal. N j 2 x x j 1 12 j 362 f ''' x j f ''' x j 1 (2) 2 This is accomplished by finding values of (2) for each interval,then adding new points in the middle of the intervals with the largest values of (2) ..\interpolation\Bli.doc Ibli.for DO I=1,NP G(I)=(4*G1(I)-G2(I))/3 GC=(9*G1(I)-G3(I))/8 GCC=(9*G1(I)-4*G3(I))/5 ERRG(I)=SQRT((G(I)-GC))**2+(G(I)-GCC)**2+(GCCGC)**2)/3) ENDDO Assignment Use the BLI in to evaluate a few test integrals between 0 and 1. Compare with analytic evaluations when possible. ..\References.htm. Plot G(x),ERRG(x), G(x)+ERRG(x), and G(x)-ERRG(x) 1. ftest = x4 2. 1 ftest x ln x Integral should be 0.886226925455 ftest x 3. ln x 1 x Integral 0 to 1 should be 1.64493406685 4. ftest x exp x 2 error integral. 291217049 1/21/2004 5. Fresnel integral. Let fun=sin((/2)x*x) . Integrate from 0 to 22 [Abramowitz and Stegun give 0.567822] This is a maximum of the integral. This is a Fresnel integral which arises in diffraction problems. It is also a chirp which arises when two stars collide. 6. 1 1 2 0.866025403785 x x 2 Integral to 1 should be 1.0471975512 ftest x 2/2